Problem: What is the area of the region bound by the graphs of $f(x)=4$, $g(x)=\dfrac{e^x}{5}$, and $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4\ln(20)-\dfrac{4e}{\ln(20)+1}+\dfrac{e}{5}$ (Choice B) B $4\ln(20) - \dfrac{19}{5} $ (Choice C) C $\dfrac{e^5}{5}-4\ln(20)$ (Choice D) D $\dfrac{e^4}{5} $
Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${2}$ ${4}$ ${6}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between $x=0$ and the point where the graphs intersect. From this we are looking to evaluate: $ \int_{0}^{b}\left( f(x)-g(x) \right)\,dx$ where $b$ is the $x$ -coordinate of the point of intersection. Finding the $x$ -coordinate of the intersection point We can find the $x$ -coordinate of the point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ 4&=\dfrac{e^x}{5}\\\\ 20 &= e^x \\\\ \ln (20) &= x \end{aligned}$ The graphs intersect where $x=\ln(20)$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $ \int_{0}^{\ln(20)}\left(4-\dfrac{e^x}{5}\right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{\ln(20)} \left( \dfrac{e^x}{5}- (4) \right) \,dx \\\\ &= 4x-\dfrac{e^x}{5}~\Bigg|_{0}^{\ln(20)} \\\\ &= \left(4\ln(20) -4 \right) - \left( \dfrac{1}{5} -0\right)\\\\ &= 4\ln(20) - \dfrac{19}{5} \end{aligned}$ Answer The area is $4\ln(20) - \dfrac{19}{5} $ square units.